Triangle ABC: Minimum value of (a^2+b^2)/c^2 | POTW #502 June 10th 2022

[anemone]

Here is this week's POTW:

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In a triangle $ABC$, it is given that $\dfrac{\cos A}{1+\sin A}=\dfrac{\sin 2B}{1+\cos 2B}$.

Find the minimum value of $\dfrac{a^2+b^2}{c^2}$.

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[anuttarasammyak]

I hope I made no errors in calculation.

Spoiler

From laws of sines
$$f:=\frac{a^2+b^2}{c^2}=\frac{\sin^2 A+\sin^2 B}{\sin^2 C}=\frac{\sin^2 A+\sin^2 B}{\sin^2 (A+B)}$$
The given condition reads
$$\frac{\cos A}{1+\sin A}=\tan B$$
By this we can delete B in the above formula to get
$$f(x)=2x-3+\frac{4}{1+x}$$
where
$$x=\sin A$$

$$f'(x)=2-\frac{4}{(1+x)^2}$$
$$f'(\sqrt{2}-1)=0$$
$$f(\sqrt{2}-1)=4\sqrt{2}-5 \approx 0.66$$ as minimum.